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| Format: | Preprint |
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2021
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| Online Access: | https://arxiv.org/abs/2105.04137 |
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| _version_ | 1866916122587037696 |
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| author | Bang-Jensen, Jørgen da Silva, Jonas Costa Ferreira Havet, Frédéric |
| author_facet | Bang-Jensen, Jørgen da Silva, Jonas Costa Ferreira Havet, Frédéric |
| contents | Let $D$ be an oriented graph. The inversion of a set $X$ of vertices in $D$ consists in reversing the direction of all arcs with both ends in $X$. The inversion number of $D$, denoted by ${\rm inv}(D)$, is the minimum number of inversions needed to make $D$ acyclic. Denoting by $τ(D)$, $τ' (D)$, and $ν(D)$ the cycle transversal number, the cycle arc-transversal number and the cycle packing number of $D$ respectively, one shows that ${\rm inv}(D) \leq τ' (D)$, ${\rm inv}(D) \leq 2τ(D)$ and there exists a function $g$ such that ${\rm inv}(D)\leq g(ν(D))$. We conjecture that for any two oriented graphs $L$ and $R$, ${\rm inv}(L\rightarrow R) ={\rm inv}(L) +{\rm inv}(R)$ where $L\rightarrow R$ is the dijoin of $L$ and $R$. This would imply that the first two inequalities are tight. We prove this conjecture when ${\rm inv}(L)\leq 1$ and ${\rm inv}(R)\leq 2$ and when ${\rm inv}(L) ={\rm inv}(R)=2$ and $L$ and $R$ are strongly connected. We also show that the function $g$ of the third inequality satisfies $g(1)\leq 4$.
We then consider the complexity of deciding whether ${\rm inv}(D)\leq k$ for a given oriented graph $D$. We show that it is NP-complete for $k=1$, which together with the above conjecture would imply that it is NP-complete for every $k$. This contrasts with a result of Belkhechine et al. which states that deciding whether ${\rm inv}(T)\leq k$ for a given tournament $T$ is polynomial-time solvable. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2105_04137 |
| institution | arXiv |
| publishDate | 2021 |
| record_format | arxiv |
| spellingShingle | On the inversion number of oriented graphs Bang-Jensen, Jørgen da Silva, Jonas Costa Ferreira Havet, Frédéric Combinatorics Discrete Mathematics Let $D$ be an oriented graph. The inversion of a set $X$ of vertices in $D$ consists in reversing the direction of all arcs with both ends in $X$. The inversion number of $D$, denoted by ${\rm inv}(D)$, is the minimum number of inversions needed to make $D$ acyclic. Denoting by $τ(D)$, $τ' (D)$, and $ν(D)$ the cycle transversal number, the cycle arc-transversal number and the cycle packing number of $D$ respectively, one shows that ${\rm inv}(D) \leq τ' (D)$, ${\rm inv}(D) \leq 2τ(D)$ and there exists a function $g$ such that ${\rm inv}(D)\leq g(ν(D))$. We conjecture that for any two oriented graphs $L$ and $R$, ${\rm inv}(L\rightarrow R) ={\rm inv}(L) +{\rm inv}(R)$ where $L\rightarrow R$ is the dijoin of $L$ and $R$. This would imply that the first two inequalities are tight. We prove this conjecture when ${\rm inv}(L)\leq 1$ and ${\rm inv}(R)\leq 2$ and when ${\rm inv}(L) ={\rm inv}(R)=2$ and $L$ and $R$ are strongly connected. We also show that the function $g$ of the third inequality satisfies $g(1)\leq 4$. We then consider the complexity of deciding whether ${\rm inv}(D)\leq k$ for a given oriented graph $D$. We show that it is NP-complete for $k=1$, which together with the above conjecture would imply that it is NP-complete for every $k$. This contrasts with a result of Belkhechine et al. which states that deciding whether ${\rm inv}(T)\leq k$ for a given tournament $T$ is polynomial-time solvable. |
| title | On the inversion number of oriented graphs |
| topic | Combinatorics Discrete Mathematics |
| url | https://arxiv.org/abs/2105.04137 |